$h(a, b) = b + \cos(ab)$ What is $\dfrac{\partial h}{\partial a}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $1 - 2a\sin(ab)$ (Choice B) B $-b\sin(ab)$ (Choice C) C $1 - a\sin(ab)$ (Choice D) D $-\sin(ab)-ab\cos(ab)$
Solution: Taking a partial derivative with respect to $a$ means treating $b$ like a constant, then taking a normal derivative. $\begin{aligned} \dfrac{\partial h}{\partial a} &= \dfrac{\partial}{\partial a} \left[ b + \cos({a}b) \right] \\ \\ &= \dfrac{\partial}{\partial a} \left[ b \right] + \dfrac{\partial}{\partial a} \left[ \cos({a}b) \right] \\ \\ &= 0 - b\sin({a}b) \end{aligned}$ In conclusion, $\dfrac{\partial h}{\partial a} = -b\sin(ab)$